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So I wish to find for each of these functions a Lipschitz constant or prove that none exists. So my definition for a function to be Lipschitz is:

A function $f:[a,b] \rightarrow \mathbb{R}$ is Lipschitz if there exists a $L$ such that $|f(x) - f(y)| \leq L|x-y|$ for all $x,y \in [a,b]$.

  1. $f(x) = \frac{1}{x}$ for $x \in (0, 1]$

  2. $f(x) = e^x$ for $x \in \mathbb{R}$

  3. $f(x) = \sqrt{1-x^2}$ for $x \in [-1,1]$

My attempt for 1) to prove $f$ is not Lipschitz is via contradiction. Suppose that $x, y \in (0,1]$ and $f$ is Lipschitz. Then there exists a $L$ such that,

$$|\frac{1}{x} -\frac{1}{y}| = \frac{|x-y|}{|xy|} \leq L |x-y|$$ for all $x,y \in (0, 1]$. This would imply that, $\frac{1}{|xy|} \leq L$ for all $x,y \in (0, 1]$. But such a $L$ cannot exist since we can make $x, y$ as small as we like, the fraction will grow.

So in conclusion I attempted 1) but not sure if I am correct. I am stuck on 2 and 3. Anybody can give me some hints? I'm thinking of using the Mean Value Theorem on question 2. Other than that I have no idea how to start.

David
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    Your attempt for 1) looks good. 2) is also not Lipschitz on $\Bbb R$ (however it is Lipschitz on any bounded interval). – Surb Aug 18 '15 at 08:09

2 Answers2

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You are right concerning 1).

Concerning 2): assume that there exist $L$ such that bla-bla-bla. Then $|\Bbb e ^ x - 1| \le L |x - 0|$, or $\frac {\Bbb e ^ x - 1} x \le L$ for $x>0$. What happens for $x \to \infty$?

For 3): note that if a continuously derivable function is Lipschitz, then $|f(x) - f(x_0)| \le L|x - x_0|$, so $|\frac {f(x) - f(x_0)} {x-x_0} | \le L$, so $|f'(x_0)| \le L$. Now show that $(\sqrt{1-x^2}) '$ is not bounded on $(-1,1)$ by studying its behaviour toward $\pm 1$, which shows that no such $L$ can be found, therefore the function is not Lipschitz.

Alex M.
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  • Cheers, we can say for 2) it will tend to infinity since an exponential 'grows' faster than the linear term. I'm not sure how to phrase this though. I'll have a crack at 3. – David Aug 18 '15 at 08:23
  • @David: To show that $\lim \limits _{x \to \infty} \frac {\Bbb e^x} x = \infty$ use L'Hospital's theorem. – Alex M. Aug 18 '15 at 08:25
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    An easier approach for 2) is to consider $y=x-1$, i.e. checking if for some $L$, $e^x-e^{x-1}\le L$ for all $x$. – Klaus Draeger Aug 18 '15 at 09:15
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You're correct for 1). Now for 2) suppose that there is $L$ such that

$$|e^x-e^y|\le L|x-y|\; \forall x,y\in \Bbb R$$ then for $x=n$ and $y=0$ we find that the sequence $\frac{e^n-1}n$ is bounded which is a contradiction. Can you prove 3) by the same way?

  • Thanks! I can see that the exponential function grows faster than the linear function. I'm not sure how to phrase it though. – David Aug 18 '15 at 08:23
  • Yes this is the way, but I think that you should tend $x$ to $\pm1$ to find the contradiction. –  Aug 18 '15 at 08:36
  • So does this mean we can select $y = 1$, keeping the variable $x$. So, we get after rationalizing, $|\frac{x+1}{\sqrt{1-x^2}}| \leq L$. So if we tend x to say 1, the bottom decreases to something very small hence the whole fraction will tend to infinity. Thus we cannot find a bound $L.$ – David Aug 18 '15 at 08:41
  • Multiplied top and bottom by $\sqrt{1-x^2}$ – David Aug 18 '15 at 08:55
  • Ah ok sorry, that's correct:-) –  Aug 18 '15 at 08:58